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80 lines
2.4 KiB
Fortran
80 lines
2.4 KiB
Fortran
*DECK SSCAL
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SUBROUTINE SSCAL (N, SA, SX, INCX)
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C***BEGIN PROLOGUE SSCAL
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C***PURPOSE Multiply a vector by a constant.
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C***LIBRARY SLATEC (BLAS)
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C***CATEGORY D1A6
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C***TYPE SINGLE PRECISION (SSCAL-S, DSCAL-D, CSCAL-C)
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C***KEYWORDS BLAS, LINEAR ALGEBRA, SCALE, VECTOR
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C***AUTHOR Lawson, C. L., (JPL)
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C Hanson, R. J., (SNLA)
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C Kincaid, D. R., (U. of Texas)
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C Krogh, F. T., (JPL)
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C***DESCRIPTION
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C
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C B L A S Subprogram
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C Description of Parameters
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C
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C --Input--
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C N number of elements in input vector(s)
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C SA single precision scale factor
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C SX single precision vector with N elements
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C INCX storage spacing between elements of SX
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C
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C --Output--
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C SX single precision result (unchanged if N .LE. 0)
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C
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C Replace single precision SX by single precision SA*SX.
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C For I = 0 to N-1, replace SX(IX+I*INCX) with SA * SX(IX+I*INCX),
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C where IX = 1 if INCX .GE. 0, else IX = 1+(1-N)*INCX.
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C
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C***REFERENCES C. L. Lawson, R. J. Hanson, D. R. Kincaid and F. T.
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C Krogh, Basic linear algebra subprograms for Fortran
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C usage, Algorithm No. 539, Transactions on Mathematical
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C Software 5, 3 (September 1979), pp. 308-323.
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C***ROUTINES CALLED (NONE)
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C***REVISION HISTORY (YYMMDD)
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C 791001 DATE WRITTEN
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C 890831 Modified array declarations. (WRB)
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C 890831 REVISION DATE from Version 3.2
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C 891214 Prologue converted to Version 4.0 format. (BAB)
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C 900821 Modified to correct problem with a negative increment.
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C (WRB)
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C 920501 Reformatted the REFERENCES section. (WRB)
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C***END PROLOGUE SSCAL
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REAL SA, SX(*)
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INTEGER I, INCX, IX, M, MP1, N
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C***FIRST EXECUTABLE STATEMENT SSCAL
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IF (N .LE. 0) RETURN
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IF (INCX .EQ. 1) GOTO 20
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C
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C Code for increment not equal to 1.
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C
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IX = 1
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IF (INCX .LT. 0) IX = (-N+1)*INCX + 1
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DO 10 I = 1,N
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SX(IX) = SA*SX(IX)
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IX = IX + INCX
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10 CONTINUE
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RETURN
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C
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C Code for increment equal to 1.
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C
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C Clean-up loop so remaining vector length is a multiple of 5.
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C
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20 M = MOD(N,5)
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IF (M .EQ. 0) GOTO 40
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DO 30 I = 1,M
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SX(I) = SA*SX(I)
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30 CONTINUE
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IF (N .LT. 5) RETURN
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40 MP1 = M + 1
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DO 50 I = MP1,N,5
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SX(I) = SA*SX(I)
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SX(I+1) = SA*SX(I+1)
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SX(I+2) = SA*SX(I+2)
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SX(I+3) = SA*SX(I+3)
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SX(I+4) = SA*SX(I+4)
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50 CONTINUE
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RETURN
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END
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